Showing the existence of an automorphism $v$ such that $v=f$ in the supplement of $Ker(f)$

Question

Let $f$ be an endomorphism of a vector space $E$ that has a finite dimension. Let $F$ be a sub vector space of $E$ such that $E=F + Ker(f)$ is a direct sum, $Ker(f)$ is the kernel of $f$.
Show that an automorphism $v$ of $E$ exists such that for all $x \in F$ : $v(x)=f(x)$.
I found this exercice in the top of a lengthy problem of Ideals in linear algebra as a fundemental result to use in the following questions. Since we have mapped all elements of $F$, I have to map elements of the Kernel and assign them values such that $v$ is an ismorphism so I thought of assigning them to a sub vector space that is in direct sum with $Im(f)$ but I can't even verify linearity...other than that I'm quite stuck on this one. Any help is appreciated!